2 Answers
2

The value of $a^3+b^3+c^3$ is not determined. Just choose $a=c$. But leaving out the condition $a\ne c$ is probably an oversight, so assume from now on that $a\ne c$.

If $q$ is a common root, then $aq^2+bq+c=cq^2+bq+a=0$. Subtracting, we find that $(a-c)q^2-(a-c)=0$. Since $a\ne c$, we get $q^2=1$.

We cannot have $q=-1$, for that implies that $b=a+c$. We are left with $q=1$, which gives $a+b+c=0$. Conversely, if $a+b+c=0$, then $1$ is a common root of the two equations.

That still leaves many possibilities for the value of $a^3+b^3+c^3$. For example, let $a=1$, $b=-3$, $c=2$. Then $a^3+b^3+c^3=-18$. Let $a=1$, $b=-4$, $c=2$. Then $a^3+b^3+c^3=-36$.

However, we can say something interesting about $a^3+b^3+c^3$ in the case $a\ne c$.
Use the general identity
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-bc-ca-ab),\tag{$1$}$$
which can be verified by multiplying out. Since in our case $a+b+c=0$, we conclude that
$$a^3+b^3+c^3=3abc.$$
Perhaps this is the intended answer.